# Symmetries in Wilsonian RG (2)

This question is related to the paper http://arxiv.org/abs/1204.5221 and is a continuation of the previous question Symmetries in Wilsonian RG

• In the liked paper why do the equalities in equation 2.7 and 2.11 hold? (the LHS of both the equations is the same and hence the two equations are 2 different ways of writing the full connected functional W)

I guess one reads 2.7 to say that when one is flowing down to the IR from UV one develops only "relevant" (dim <4) operators and one I guess reads 2.11 to mean that one develops only irrelevant (dim >4) operators when one flows up to the UV from IR.

Why?

• In the linked paper just below equation 2.2 the authors comment that if there is a CFT in the UV then this UV behaviour can change if irrelevant operators are added. why? I would think that (dim>4)/irrelevant operators would come suppressed with positive powers of the cut-off and hence if one pushes the cut-off to infinity then they would vanish and hence the UV is not affected by them. But the authors don't seem to think so...

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@useer1504 Can you put in your definitions? If O is an irrelevant operator of dimension $4 + n >4$ then would flow in the IR as $O/\Lambda ^n$ where $\Lambda$ is the UV cut-off. Hence in the IR such a high dimension operator will become important since it scales inversely with $\Lambda$. What is wrong in this argument? – user6818 Apr 11 '13 at 22:48
@user6818 $\Lambda$ is a UV momentum cutoff. The IR limit corresponds to $\Lambda \to \infty$. – user1504 Apr 12 '13 at 13:41
You're computing the expectation value $\langle \mathcal{O} \rangle$ of an observable which has some characteristic energy scale $\lambda$ (equivalently, a characteristic distance scale $\hbar/\lambda$). The contribution to this observable from an irrelevant operator of dim $4 + n$ will be of order $(\lambda/\Lambda)^n$. Take the IR limit means taking the energy scale of the observable to be very small relative to the energy scale $\Lambda$ of the cutoff. – user1504 Apr 12 '13 at 23:50
@user6818 It occurs to me that the customary notation $\Lambda \to \infty$ is very sloppy because it obscures the critical fact that two distance scales are being compared when one approaches an IR or UV limit. – user1504 Apr 12 '13 at 23:56